Bandwidth in GFSK

Hello, I have a doubt how data rate is calculated in GFSK ? I am new to RF. so please describe.my second question is why can"t we change frequency faster for high data rate? a typical 433Mhz transceiver hardly gives 100Kbits only why?

Welcome to PF.
The FSK data must pass through the very narrow 433MHz channel allocation without causing co-channel interference. The Gaussian filter is probably being used to remove the broadband “key clicks” generated by the bit transitions.

Yes it can, but with only a 1Hz difference you would have to sample for one entire second before you knew for sure which it was.
Data rate is proportional to the bandwidth.
1Hz BW is about 1 bit per second.
100 kHz BW is about 100 kbps.

First of all ,thanks for your immediate response and sharing your wisdom. in the above post you said i have to wait for 1 second to detect the frequency I understand that but is this possible to measure the wavelength so that even a single wave will give the frequency & i will not have to wait for that long.

433 MHz= 433 waves in 1 us===> 216.5 waves in 0.5 us.what if i take those 216 waves and calculate the time period from them hence frequency ? if it's impossible why? if possible it would give a huge data rate. since 0.5us for each bit then 2bits in 1us=>2Kbits in 1ms=>2Mbits in 1 second.it's still far better than 100kbits. Note these are my theoretical assumptions .

433 MHz= 433 waves in 1 us===> 216.5 waves in 0.5 us.what if i take those 216 waves and calculate the time period from them hence frequency ? if it's impossible why? if possible it would give a huge data rate. since 0.5us for each bit then 2bits in 1us=>2Kbits in 1ms=>2Mbits in 1 second.it's still far better than 100kbits. Note these are my theoretical assumptions.

Fundamentally, a radio channel has a bandwidth just like a low-pass filter has a cut-off frequency. The data rate through a low pass filter or radio channel cannot be greater than the actual bandwidth of that channel.

To measure the frequency of a 1 MHz sinewave to one part in 1 thousand requires a 1 MHz * 1kHz = 1 GHz clock. Now a nice clean sinewave has a measurable frequency, but when you add other noise to that sinewave it is no longer possible to identify where the critical zero crossings are to within 1 nanosecond. To find those zero crossings you must fit a sine wave to the received signal, that will take one whole cycle per measurement. Detection is multiplication which is correlation of the received signal with a reference waveform.

One bit every 0.5us = 2 M bits per second, which requires a 2 MHz bandwidth. Unfortunately the channels allocated on the 433MHz band are only 25 kHz apart, spread over 1.7 MHz. What happens to the other 68 users when your data interferes with their low data rate signals. http://en.wikipedia.org/wiki/LPD433

If you want higher data rates then you need more bandwidth. That requires a microwave link or an optic fibre. There is a big difference between the low data rate links between distributed microcontrollers and an 11 Mbit/sec datalink on 2.45GHz.

These days, everything to do with modulation seems to have become digital signal processing, so that is now the best approach. Your experience, field of interest and mathematical ability will dictate the best reference.